1
$\begingroup$

In the context of SM ($SU(3)_C\otimes SU(2)_L\otimes U(1)_Y$) the charge operator is $Q_{SM} = T_3 + \frac{Y}{2}\mathbb{I}_2$ and gives us the fermions charges. Here $T_3=\frac{1}{2}\sigma_3$ is the third $SU(2)$ generator.

For example, assuming $Y=-1$ for the left lepton doublet, $Q_{SM}\Psi=\begin{pmatrix}0&0\\0&-1\end{pmatrix}\begin{pmatrix}\nu_L \\ e_L\end{pmatrix}=\begin{pmatrix}0 \nu_L \\ -1 e_L\end{pmatrix}$ and the charge components are obtained.

This process works fine for the scalar doublet too, $Q_{SM}\begin{pmatrix}\phi^+\\\phi^0\end{pmatrix}=\begin{pmatrix}+1 \phi^+\\0 \phi^0\end{pmatrix}$, with $Y=+1$.

On the other hand, the same calculation can be applied for $SU(3)_C\otimes SU(3)_L\otimes U(1)_X$ extension in which the charge operator is $Q_{331} = T_3-\sqrt{3}T_8+X\mathbb{I}_3$ where $T_3=\frac{1}{2}\lambda_3$ and $T_8=\frac{1}{2}\lambda_8$ are the $SU(3)$ diagonal generators.

For example, $Q_{331}\Psi_1=\begin{pmatrix}0&0&0\\0&-1&0\\0&0&1\end{pmatrix}\begin{pmatrix}\nu_L \\ e_L \\e_L^c\end{pmatrix}=\begin{pmatrix}0 \nu_L \\ -1 e_L \\ +1 e_L^c\end{pmatrix}$ and the charge components are obtained again with $X=0$.

In the same way as before, we can do this with the scalar multiplets. For example with one of the triplets called $\eta$, $Q_{331}\eta=\begin{pmatrix}1&0&0\\0&0&0\\0&0&2\end{pmatrix}\begin{pmatrix}\eta^+ \\ \eta^0 \\ \eta^{++}\end{pmatrix}=\begin{pmatrix}+1 \eta^+ \\ 0 \eta^0 \\ +2 \eta^{++}\end{pmatrix}$ and the charge components are obtained again with $X=+1$.

The question is, ¿what about the matrix scalar multiplets? e.g. for a 331 sextet defined as $S=\begin{pmatrix} \sigma_1^0&h_2^-&h_1^+ \\ h_2^- & H_1^{--} & \sigma_2^0 \\ h_1^+ & \sigma_2^0 & H_2^{++}\end{pmatrix},$ with hypercharge $X=0$.

When you try to do the same process, but with $Q_{331}^{\dagger}SQ_{331}$ for being $S$ a matrix and not a vector, i can't obtain the charges. Here $Q_{331}$ is the same as for $\Psi_1$.

¿Any idea in order to obtain the component charges?

$\endgroup$
  • $\begingroup$ What is the representation of the sextet $S$ under the gauge group? $\endgroup$ – InertialObserver Feb 21 at 18:04
  • $\begingroup$ In the group $SU(3)_C\otimes SU(3)_L\otimes U(1)_X$ which contains SM group, the sextet transforms as $S\sim (1_C,6_L,0)$. That is $S$ is a singlet in $SU(3)_C$, a sextet in $SU(3)_L$ with hypercharge zero. $\endgroup$ – Eduardo Castillo Ruiz Feb 21 at 23:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.